Calculate the volume of $\iiint (x^2+y^2+z^2)^2\,dB$ while $B$ is a sphere with radius $2$ with its center in the origin.
I have the solution in the book, they did it the obvious way using sphere coordinates which is something I know how to do but I am simply trying to learn as many ways as possible.
what I tried is $B: x^2+y^2+z^2=4$ then divided by $4$ to try and get the sphere with radius $1$ and set $u= \frac{x}{2}$ $v= \frac{y}{2}$ $w= \frac{z}{2}$ and the jacobian would be $J^{-1}=8$ so now the integral looks like $8 \cdot 4 \iiint(u^2+v^2+w^2)^2\,dV$ but I keep getting wrong answer in the ways I try from here for example if now I move to sphere coordinates and calculate $$\iiint r^6 \cdot \sin \phi\, dr\,d\phi \,d\theta\,,\text{ while }\quad 0 \leq r \leq 1\,,\quad 0 \leq \phi \leq \pi\,, \quad 0 \leq \theta \leq 2 \pi$$ I get a wrong answer.
if I put $u^2+v^2+w^2=1$ and then just calculate the sphere volume using $\frac{4}{3} \pi r^3$ I also get the wrong answer.
how can I make this way work if it is possible(although it does not really help in this case but just practicing the change of variables and such..)? the correct anwer is $\frac{512 \pi}{ 7}\,.$
Your mistake is that you think, that factor before integral is $8\cdot 4$, while it should be $8\cdot 2^4 = 2^7$:
$8$ comes, as you correctly calculated, from jacobian and $2^4$ from $(x^2+y^2+z^2)^2 = 2^4 (u^2+v^2+w^2)^2$
After changing variables your integral is $\frac{4\pi}{7}$, so difference between initial one and last exactly is $128=2^7$ and we can be calm about the operation of variable substitution - it works.